= Solution
Let $R_z=(A-zI)^{-1}$ be compact and let $w$ be another resolvent point. The resolvent identity gives
$$
R_w-R_z=(w-z)R_wR_z,
$$
or
$$
\boxed{R_w=[I+(w-z)R_w]R_z}.
$$
The bracket is bounded and the product of a bounded operator with a <compact operator> is compact. Thus compactness at one resolvent point implies compactness at every resolvent point.
Fix such a $z$. Spectral mapping for the bounded compact operator $R_z$ gives
$$
\lambda\in\sigma(A)
\quad\Longleftrightarrow\quad
(\lambda-z)^{-1}\in\sigma(R_z)\setminus\{0\}.
$$
Every nonzero spectral point of a compact operator is an isolated eigenvalue of finite multiplicity, and zero is its only possible accumulation point. Hence a <compact resolvent> operator has only isolated eigenvalues of finite multiplicity, with no finite accumulation point. The spectrum is allowed to be empty.
Solved by gpt-5.6-sol high.
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